Question about not isomorphic mapping between affine varieties I try to understand the following example: $ F :\mathbb{A}^1 \to \mathbb{A}^2 , t \mapsto (t^2,t^3)$. And let $X=\mathbb{A}^1$, $Y= \operatorname{Im}F=\left \{ (x,y) : F(t)=(x,y) \right \}$ (In fact, $Y=Z((y^2=x^3)) \subset \mathbb{A}^2$) Then, $F$ is NOT isomorphism between $X$ and $Y$(although $F$ is a homeomoprhism)
To begin with, the most important idea to check the following statement is here:

Theorem. Let $X \subset \mathbb{A}^n, Y \subset \mathbb{A}^m $ be affine varieties, then there is a one-to-one correspondence between
$\left \{ F: X \to Y ~ is ~ a ~morphism \right \} \longleftrightarrow
 \left \{ F^{*} : A(Y) \to A(X) ~ is ~ a ~k-algebra~ homomorphism 
 \right \} $

So, in order to verify the statement, suppose $F$ is an isomorphism. Then since by Theorem,
naturally, there exists an $k$-algebra isomorphism $F^{*} : A(Y) \to A(X)$. Actually, $A(Y)=k[x,y]/(y^2-x^3), A(X)=k[x]$. Then, the contradiction occurs when  $F^{*}(\mathcal{O}_Y(V)) \nsubseteq  \mathcal{O}_{X}(F^{-1}(V))$ for some open set $V$  in $Y$, i.e  $F$  is NOT morphism. Or, when considering the T.F.A.E of the definition of morphism,

Claim : There exists a map $\psi \in A(Y) $ such that $F^{*}\psi \notin A(X) $.

However, I do not understand how construct the map $F$,
$F^*(x,y)(t):= (F^{*}(x), F^{*}(y))(t)=(t^2,t^3)$.
Then, since the image of $F^* $ is not an element of $A(X)(=k[x])$, it is contradictory that $F$ is morphism(hence, naturally $F$ cannot be isomorphism). But I cannot understand why the map $F^*$ is defined like the above.
To be specific, when I think about the construction of the map $F^*$, first pick $\psi \in A(Y)$. Then  since $A(Y)=k[x,y]/(y^2-x^3)$, $\psi$ is  two-variable polynomial $x,y$, so $\psi= \psi(x,y)$
So, in my opinion, $\psi \mapsto F^{*}(\psi(x,y))=★ \in A(X) $ seems to be natural. But my idea is a totally different form when comparing to the above. And how the variable $t$ come?
 A: More generally, let $X \subset \mathbb{A}^n$ and $Y \subset \mathbb{A}^m$ be two affine varieties, defined by $(\varphi_1,...,\varphi_s)$ and $(\psi_1,...,\psi_s)$, $f$ be a morphism from $X$ to $Y$ defined by $(f_1,...,f_m)$. You can think this m $f_i$ are just polynomials in indeterminants $(x_1,...,x_n)$ satisfying certain conditions (if $(a_1,...,a_n)\in\mathbb{A}^n$ satisfy the functions $(\varphi_1,...,\varphi_s)$ then $(f_1(a_1,...,a_n),...,f_m(a_1,...,a_n))\in\mathbb{A}^m$ satisfy the functions $(\psi_1,...,\psi_s)$, because they together define the morphism $f$ from $X$ to $Y$).
The morphism $f^*$ from the coordinate ring of $Y$ $A(Y)=k[y_1,...,y_m]/(\psi_1,...,\psi_r)$ to the coordinate ring of $X$ $A(X)=k[x_1,...,x_n]/(\varphi_1,...,\varphi_s)$ is just the pull back (or composition with $f$)of 'functions on $Y$' to 'functions on $X$'(here the functions on $X$ or $Y$ means the morphism from $X$ or $Y$ to $\mathbb{A}^1$ which is also elements in their coordinate rings).Thinking $y_i\in A(Y)$ is a function on $Y$, so its pull bcak $f^*(y_i)\in A(X)$ is its composition with $f$: $f^*(y_i)(x_1,...x_n) = (y_i\circ f) (x_1,...x_n) = y_i(f_1(x_1,...,x_n),...,f_m(x_1,...,x_n)) = f_i(x_1,...x_n)$. \ You can check this indeed defines a morphism from $A(Y)$ to $A(X)$ (just check the elements in the ideal $(\psi_1,...,\psi_r)$ is mapped into the ideal $(\varphi_1,...,\varphi_s)$, but it's the same thing with the conditions on the $f_i$'s mentioned above).
A: Two affine varieties $X:=V(I)\subseteq \mathbb{A}^m_k, Y:=V(J) \subseteq \mathbb{A}^n_k$ are isomorphic (as varieties over the algebraically closed field $k$) iff their coordinate rings $A(X),A(Y)$ are isomorphic as $k$-algebras. Hence to check that they are not isomorphic you must prove there is no isomorphism $\phi: A(X) \cong A(Y)$ of $k$-algebras.
Note: In general if given two commutative rings (or $k$-algebras) $A,B$ and you ask: "Are these two rings isomorphic (as $k$-algebras)?", this question - even though it is an "elementary" question, may not be easy to answer.
In this case the ring $k[x]$ is regular and the ring $k[x,y]/(y^2-x^3)$ is not regular and regularity is preserved under isomorphism,  hence $X\ncong Y$. An isomorphism between varieties $f:X \cong Y$ induce an isomorphism between their tangent spaces $T_x(X) \cong T_{f(x)}(Y)$, and in this case it follows the tangent space of $Y$ at $y:=(0,0)$ is 2-dimensional. The tangent space of $X$ at $x:=(0)$ is one dimensional. The induced map
$$df_0:T_0(X) \rightarrow T_0(Y)$$
is not an isomorphism. Hence the map $F$ you write down cannot be an isomorphism.
A: You can see that this is not an isomorphism geometrically: indeed, the differential of $F$ is zero at the origin, so that the map cannot be an isomorphism.
